Two solids $$A$$ and $$B$$ of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies $$(K.E.)_A : (K.E.)_B$$ will be $$\frac{A}{1}$$, so the value of $$A$$ will be ______.

The kinetic energy of a body can be expressed in terms of its linear momentum $$p$$ as $$K.E. = \frac{p^2}{2m}$$, since $$p = mv$$ gives $$K.E. = \frac{1}{2}mv^2 = \frac{p^2}{2m}$$.

Given that solids $$A$$ and $$B$$ have equal linear momentum, i.e., $$p_A = p_B = p$$, their kinetic energies are $$(K.E.)_A = \frac{p^2}{2m_A} = \frac{p^2}{2 \times 1} = \frac{p^2}{2}$$ and $$(K.E.)_B = \frac{p^2}{2m_B} = \frac{p^2}{2 \times 2} = \frac{p^2}{4}$$.

The ratio of their kinetic energies is $$\frac{(K.E.)_A}{(K.E.)_B} = \frac{p^2/2}{p^2/4} = \frac{4}{2} = \frac{2}{1}$$.

Since the ratio is given as $$\frac{A}{1}$$, the value of $$A$$ is $$2$$.

The correct answer is 2.